Sodium hypochlorite, NaClO is the active ingredient in many bleaches. Calculate the ratio of the concentration of `ClO^-` and HClO in a bleach having a pH adjusted to 6.50 by using strong acid and strong base.

Explanation

To solve the problem of calculating the ratio of the concentrations of \( \text{ClO}^- \) and \( \text{HClO} \) in a bleach solution with a pH of 6.50, we can follow these steps:

### Step 1: Determine the \( K_a \) and \( pK_a \) of \( \text{HClO} \)
The dissociation constant \( K_a \) for hypochlorous acid (\( \text{HClO} \)) is given as \( 3.0 \times 10^{-8} \).

To find \( pK_a \):
\[ pK_a = -\log(K_a) = -\log(3.0 \times 10^{-8}) \]
Calculating this gives:
\[ pK_a \approx 7.523 \]

### Step 2: Use the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation relates pH, \( pK_a \), and the ratio of the concentrations of the conjugate base and the weak acid:
\[ \text{pH} = pK_a + \log\left(\frac{[\text{ClO}^-]}{[\text{HClO}]}\right) \]

Substituting the known values:
\[ 6.50 = 7.523 + \log\left(\frac{[\text{ClO}^-]}{[\text{HClO}]}\right) \]

### Step 3: Rearranging the Equation
Rearranging the equation to isolate the logarithmic term:
\[ \log\left(\frac{[\text{ClO}^-]}{[\text{HClO}]}\right) = 6.50 - 7.523 \]
Calculating the right side:
\[ \log\left(\frac{[\text{ClO}^-]}{[\text{HClO}]}\right) = -1.023 \]

### Step 4: Exponentiating to Find the Ratio
To eliminate the logarithm, we exponentiate both sides:
\[ \frac{[\text{ClO}^-]}{[\text{HClO}]} = 10^{-1.023} \]
Calculating this gives:
\[ \frac{[\text{ClO}^-]}{[\text{HClO}]} \approx 0.0948 \]

### Step 5: Expressing the Ratio
This means the ratio of \( [\text{ClO}^-] \) to \( [\text{HClO}] \) can be expressed as:
\[ \frac{[\text{ClO}^-]}{[\text{HClO}]} \approx 0.0948 : 1 \]

### Final Answer
Thus, the ratio of the concentration of \( \text{ClO}^- \) to \( \text{HClO} \) in the bleach solution is approximately:
\[ 0.0948 : 1 \]

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